Prove that cot A + cot (60 + A) –  cot (60 – A) = 3 cot 3A.

Solution :

We have,

L.H.S = cot A + cot (60 + A) –  cot (60 – A)

\(\implies\) L.H.S = \(1\over tan A\) + \(1\over tan (60 + A)\) – \(1\over tan (60 – A)\)

\(\implies\) L.H.S = \(1\over tan A\)+ \(1 – \sqrt{3} tan A\over \sqrt{3} + tan A\) – \(1 + \sqrt{3} tan A\over \sqrt{3} – tan A\)

[ By using this formula, tan (A + B) = \(tan A + tan B\over 1 – tan A tan B\)  above ]

\(\implies\) L.H.S = \(1\over tan A\) – \(8 tan A\over 3 –  tan^2 A\)

\(\implies\) L.H.S = \(3 – 9 tan^2 A\over 3 tan A – tan^3 A\)

L.H.S = 3(\(1 – 3 tan^2 A\over 3 tan A – tan^3 A\))

Since \(3 tan A – tan^3 A\over 1 – 3 tan^2 A\) = tan 3A

\(\implies\) L.H.S = \(3\over tan 3A\) = 3 cot 3A = R.H.S

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